**Fundamental Principle of Counting**

“If an event can occur in m different ways, following which another event can occur in n different ways, then the total number of occurrence of the events in the given order is m × n.”

The above principle can be generalized for any finite number of events.

“If an event can occur in m different ways, following which another event can occur in n different ways, following which a third event can occur in p different ways, then the total number of occurrence to ‘the events in the given order is m × n × p.”

KeywordsAND➜ MULTIPLICATION (Product rule)

OR➜ ADDITION (Sum Rule)

**What is Permutation? **

A permutation is defined as an arrangement in a definite order of a number of objects taken, some or all at a time. Counting permutations are merely counting the number of ways in which some or all objects at a time are rearranged. The convenient expression to denote permutation is defined as “ ^{n}P_{r }”.

The permutation formula is given by:

Pr = n!/(n-r)! ; 0 ≤ r ≤ nWhere the symbol “!” denotes the factorial which means that the product of all the integers is less than or equal to n but it should be greater than or equal to 1.

**Factorial Notation:**

The notation n! represents the product of the first n natural numbers, i.e., the product 1 × 2 × 3 × . . . × (n – 1) × n is denoted as n!. We read this symbol as ‘n factorial’.

Thus, 1 × 2 × 3 × 4 . . . × (n – 1) × n = n !

For example,

1! = 1

2! = 1 x 2 = 2

3! = 1 x 2 x 3 = 6

4! = 1 x 2 x 3 x 4 = 24, which are the factors of the given number.

**Permutation When all the Objects are Distinct: **

There are some theorems involved in finding the permutations when all the objects are distinct. They are :

**Theorem 1: **If the number of permutations of n different objects taken r at a time, it will satisfy the condition 0 < r ≤ n and the objects which do not repeat is n ( n – 1) ( n – 2)……( n – r + 1), then the notation to denote the permutation is given by “ ^{n }P_{r}”

**Theorem 2: T**he number of permutations of different objects “n” taken r at a time, where repetition is allowed and is given by n^{r}.

**Permutation When all the Objects are not Distinct Objects: **

**Theorem 3**: To find the number of permutations of the objects ‘n’, and ‘p’s are of the objects of the same kind and the rest is all different is given as n! / p!

**Theorem 4**: The number of permutations of n objects, where p_{1} are the objects of one kind, p_{2} are of the second kind, …, p_{k }is of the k^{th} kind and the rest, if any, are of a different kind, then the permutation is given by n! / ( p_{1}!p_{2}!…P_{k}!)

**What is Combination?**

The combination is a selection of a part of a set of objects or a selection of all objects when the order doesn’t matter. Therefore, the number of combinations of n objects taken r at a time and the combination formula is given by: **nCr = n(n-1)(n-2)…(n-r+1)/ r! = n!/ r!(n-r)!= nPr /r!**

**Relationship Between Permutation and combination:**

In permutation and combination for class 11, the relationship between the two concepts is given by two theorems. They are;

**Theorem 5**: ^{n}P_{r} = ^{n}C_{r }r! ; if 0 < r ≤ n.

**Theorem 6: ^{n}**C

_{r }+

^{n}C

_{r-1}=

^{n+1}C

_{r}

**Try yourself:****Find the number of words, with or without meaning, that can be formed with the letters of the word ‘INDIA’.**

**a.**60**b.**65**c.**70**d.**75

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**Solved Questions**

**Q.1. From a pack of cards in how many ways can you select-**

**a) A king and a queen**

**b) A king or a queen**

There are 52 cards in a pack, with 4 kings and 4 queens.

A king can be selected in 4 ways and a queen can be selected in 4 ways. Number of ways of selection of a king AND a queen is 4×4=16 (Product rule)

Number of ways of selection of a king OR a queen 4+4=8 (Addition rule)

The fundamental principle of counting (FPC) is the basic concept of permutation and combination.

**Q.2. ****In how many ways can 3 people A, B, C be arranged in 2 places?**

Manually solving we will get 6 cases as given below.

AB BA

AC CA

BC CB

This can be explained using FPC like this

There are two seats __ __. In the first seat A,B or C can sit. i.e., 3 possibilities.

Let A sit in the first place. Thus, in second place either B or C can sit.

The second seat can hence be taken up in 2 ways Together A,B ‘AND’ C can be seated in 3 x 2 = 6 ways.

Thus, the number of arrangement of 3 things at 2 places can be done in 3×2 ways.

**Q.3. In how many ways can 4 people be seated in 3 chairs?**

First seat can be filled in 4 ways.

Second seat can be filled in 3 ways.

Third seat can be filled in 2 ways.

So all together 4 people can be arranged in 3 seats in 4x3x2 = 24 ways.

Generalizing; arrangement of n things at r places can be done in [n(n-1)(n-2)…(n-r+1)] ways.

This arrangement of n things at r places is denoted by

^{n}P_{r}.

∴^{n}P_{r}= n(n-1)(n-2)…(n-r+1) = n!/(n-r)!It should be noted that in all these examples, the order of arrangement is important.

In other words, AB and BA are counted as 2 different cases. An arrangement where the order is important is called permutation

Now let us see one case where the order is not important.

**Try yourself:****In how many different ways can the letters of the word ‘LEADING’ be arranged in such a way that the vowels always come together?**

**a.**360**b.**480**c.**720**d.**5040

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**Q.4. How many doubles teams can be selected from 3 persons A, B, C?**

Let’s take a doubles team AB. Change the order, i.e. to BA. Notice that both the teams are one and the same.

So here the order is not important. Hence, it’s a case of a combination. Here one can select the team in 3 ways

AB

AC

BC

If the order is important, then the total number of arrangements can be done in

^{ n}P_{r}=^{3}P_{2}ways.

We will get the 6 cases as:AB BA

AC CA

BC CB

But in this particular question every 2! (because of A,B can be arranged in 2! Ways) cases will be counted as only one

case, i.e AB and BA is one and the same team.

So the number of selections of 2 things from 3 things can be done in =

^{ }(^{3}P_{2}) / 2!Similarly, the number of selections of 3 things from 4 things can be done in = (

^{4}P_{3}) / 3!Generalizing: the number of selections of r things from n things can be done in = (

^{n}P_{r}) / r! ways.The number of selections of r things from n different things is denoted by

^{n}C_{r},which is what we learn as “

Combination”.

∴ n_{Cr}= n_{Pr}/r!=n!/[(n-r)! r!]Thus, you can understand now, that wherever the arrangement is necessary apart from selection,

we used Permutation and when we just need to select without arranging, we use Combination.

To make the point even clearer.

**Try yourself:****Find the number of words, with or without meaning, that can be formed with the letters of the word ‘CHAIR’.**

**a.**110**b.**120**c.**130**d.**140

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**Q.5. In how many ways can you select a President and a Vice-President from 3 people?**

Is it a case of permutation or combination?

In the above question, since the word “select” appears, it may tempt you to consider combination.

So let’s take a case where A is the President and B is the Vice-President. Now change the order,

i.e. B is President and A Vice-President. Are both the cases same? No, they are not! We see that both the cases are different.

The order is important. Hence it is a case of permutation.

**Try yourself:****Among a set of 5 black balls and 3 red balls, how many selections of 5 balls can be made such that at least 3 of them are black balls.**

**a.**45**b.**46**c.**47**d.**48

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**Try yourself:****Solve the following.i) ^{30 }P _{2 }ii) ^{30 }C _{2}**

**a.**870, 435**b.**435, 870**c.**870, 470**d.**435, 835

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